Grade 5 math worksheet featuring exercises on finding the mean, median, mode, and range, including problems with missing data values.

Let’s solve each problem step by step. We’ll go box by box, left to right, top to bottom.

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Box 1: Find the median of 7, 12, 16, 12, 6, 4, 1

Step 1: Put numbers in order from smallest to largest:
→ 1, 4, 6, 7, 12, 12, 16

Step 2: Count how many numbers → 7 numbers (odd)

Step 3: Median is the middle number → 4th number = 7

✔ Answer: 7

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Box 2: Mean, median, mode, range of pencil lengths: 4, 3, 8, 3, 6, 2, 9

First, sort them: 2, 3, 3, 4, 6, 8, 9

- Mean: Add all → 2+3+3+4+6+8+9 = 35; divide by 7 → 35 ÷ 7 = 5
- Median: Middle of 7 numbers → 4th number = 4
- Mode: Most frequent → 3 (appears twice)
- Range: Largest - Smallest → 9 - 2 = 7

✔ Answers:
Mean: 5, Median: 4, Mode: 3, Range: 7

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Box 3: Mean, median, mode, range of 6, 1, 3, 1, 5

Sort: 1, 1, 3, 5, 6

- Mean: 1+1+3+5+6 = 16; 16 ÷ 5 = 3.2
- Median: Middle of 5 → 3rd number = 3
- Mode: 1 (appears twice)
- Range: 6 - 1 = 5

✔ Answers:
Mean: 3.2, Median: 3, Mode: 1, Range: 5

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Box 4: What must be missing piece so mean is 10? Data: 14, __, 9, 12

We have 4 numbers total. Let missing number = x

Mean = (sum) / 4 = 10 → sum = 40

Current sum without x: 14 + 9 + 12 = 35

So x = 40 - 35 = 5

✔ Answer: 5

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Box 5: Missing piece so mode is 4? Data: 6, 2, 3, 3, 2, 4, __

Current counts:
- 2 → appears twice
- 3 → appears twice
- 4 → appears once
- 6 → once

To make mode = 4, we need 4 to appear more than any other number.

Right now, 2 and 3 are tied at 2 times. So if we add another 4, then 4 appears twice — still tied with 2 and 3.

But wait — if we add another 4, then 4 appears twice, same as 2 and 3 → no unique mode.

Actually, to make 4 the *only* mode, it needs to appear more than twice.

Wait — let’s check again.

Current data: 6, 2, 3, 3, 2, 4 → that’s 6 numbers.

Counts:
- 2: 2 times
- 3: 2 times
- 4: 1 time
- 6: 1 time

If we add 4, then 4 becomes 2 times → still tied with 2 and 3 → no single mode.

But maybe the question allows multiple modes? Usually “the mode” implies one.

Wait — perhaps I miscounted.

Original list: 6, 2, 3, 3, 2, 4 → that’s six numbers.

Add one more → seven numbers.

If we add 4, then:

- 2: 2
- 3: 2
- 4: 2 → three-way tie → technically multimodal, but often questions expect a single mode.

Hmm. Maybe they want 4 to be the *most frequent*, even if tied? But usually not.

Wait — what if we add another 4? Then 4 appears 3 times? No — currently only one 4. Adding one makes two.

Unless... did I misread?

List: 6, 2, 3, 3, 2, 4 → yes, only one 4.

To make 4 the mode, we need it to appear more than 2 times → so we’d need to add two 4s? But there’s only one blank.

That can’t be.

Wait — perhaps I made a mistake.

Maybe the current mode isn't defined yet? Or maybe adding one 4 makes it tied, and that’s acceptable?

But let’s think differently.

What if we add 4? Then frequencies:

- 2: 2
- 3: 2
- 4: 2 → all tied → no unique mode.

But if we add something else, say 2 or 3, then that number becomes mode.

The only way for 4 to become the sole mode is if it appears 3 times — which requires adding two 4s — impossible with one blank.

Wait — unless... did I miscount the original?

Look again: "6, 2, 3, 3, 2, 4" — that’s six numbers.

Perhaps the problem expects us to add 4, making it appear twice, and since others also appear twice, it's still considered a mode? But typically, if multiple values tie, they are all modes.

But the question says “to have a mode of 4” — implying 4 should be the mode, possibly among others? Or uniquely?

This is ambiguous.

Wait — let me check standard interpretation.

In many school contexts, if you add 4, and now 2,3,4 all appear twice, then there are three modes — so 4 is *a* mode, but not *the* mode.

But the question says “have a mode of 4” — which might mean 4 should be included in the mode(s).

Alternatively, perhaps I missed something.

Another idea: maybe the list is 6,2,3,3,2,4 — and we add one number. If we add 4, then 4 appears twice, same as 2 and 3 — so modes are 2,3,4. So 4 is a mode.

If the question accepts that, then answer is 4.

If it requires 4 to be the only mode, then impossible with one addition.

Given this is likely elementary level, probably they accept 4 as the answer, assuming that adding 4 makes it a mode (even if tied).

Moreover, looking at similar problems, often they allow tied modes.

So I’ll go with 4.

✔ Answer: 4

*(Note: Strictly speaking, after adding 4, there are three modes: 2,3,4. But since the question asks for “a mode of 4”, and 4 is now a mode, it should be acceptable.)*

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Box 6: Missing piece so mean is 6? Data: 7, 12, __, 3, 11

Five numbers total. Let missing = x

Sum needed for mean 6: 6 × 5 = 30

Current sum: 7 + 12 + 3 + 11 = 33

Wait — 7+12=19, +3=22, +11=33

Then x = 30 - 33 = -3

Negative? That seems odd, but mathematically correct.

Check: 7,12,-3,3,11 → sum = 7+12=19, -3=16, +3=19, +11=30 → mean=6. Yes.

So answer is -3

✔ Answer: -3

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Box 7: Missing piece so mean is 40? Data: 20, 30, 50, 40, __

Five numbers. Sum needed: 40 × 5 = 200

Current sum: 20+30+50+40 = 140

Missing = 200 - 140 = 60

✔ Answer: 60

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Box 8: Missing piece so mean is 15? Data: 18, 11, 7, 15, 32, 27, __

Seven numbers total. Sum needed: 15 × 7 = 105

Current sum: 18+11=29, +7=36, +15=51, +32=83, +27=110

Wait — 18+11=29, +7=36, +15=51, +32=83, +27=110

But 110 is already over 105? That can’t be.

Sum needed is 105, but current sum is 110? Then missing number would be negative?

Let me recalculate:

18 + 11 = 29
29 + 7 = 36
36 + 15 = 51
51 + 32 = 83
83 + 27 = 110

Yes, 110.

Sum needed for mean 15 over 7 numbers: 15×7=105

So missing number = 105 - 110 = -5

Again, negative, but mathematically correct.

Check: 18,11,7,15,32,27,-5 → sum=18+11=29, +7=36, +15=51, +32=83, +27=110, -5=105 → mean=15. Correct.

✔ Answer: -5

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Box 9: Missing piece so mean is 23? Data: 51, 52, __, 23, 21, 17, 37, 45

Eight numbers total. Sum needed: 23 × 8 = 184

Current sum: 51+52=103, +23=126, +21=147, +17=164, +37=201, +45=246

Wait — that’s without the missing number.

List given: 51, 52, __, 23, 21, 17, 37, 45 → that’s 8 positions, one missing.

So current known numbers: 51,52,23,21,17,37,45 → seven numbers.

Sum these: 51+52=103, +23=126, +21=147, +17=164, +37=201, +45=246

Sum needed: 23×8=184

But 246 > 184? That means missing number = 184 - 246 = -62

Again negative.

Check: if we add -62, total sum = 246 - 62 = 184, mean=23. Correct.

So answer is -62

✔ Answer: -62

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Now, compiling all answers in order as per the worksheet layout:

Top row:
- Box 1: median of 7,12,16,12,6,4,1 → 7
- Box 2: mean,median,mode,range of 4,3,8,3,6,2,9 → 5, 4, 3, 7
- Box 3: mean,median,mode,range of 6,1,3,1,5 → 3.2, 3, 1, 5

Middle row:
- Box 4: missing for mean 10 in 14,__,9,12 → 5
- Box 5: missing for mode 4 in 6,2,3,3,2,4,__ → 4
- Box 6: missing for mean 6 in 7,12,__,3,11 → -3

Bottom row:
- Box 7: missing for mean 40 in 20,30,50,40,__ → 60
- Box 8: missing for mean 15 in 18,11,7,15,32,27,__ → -5
- Box 9: missing for mean 23 in 51,52,__,23,21,17,37,45 → -62

Final Answer:

Box 1: 7
Box 2: Mean=5, Median=4, Mode=3, Range=7
Box 3: Mean=3.2, Median=3, Mode=1, Range=5
Box 4: 5
Box 5: 4
Box 6: -3
Box 7: 60
Box 8: -5
Box 9: -62